Integrate #\int_{ln(2)}^{ln(8)} \frac{e^x}{1+e^x}#?

Question

#\int_{ln(2)}^{ln(8)} \frac{e^x}{1+e^x}dx#

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Answer 1 · 2017-12-11T17:35:44

The answer is #=1.099#

Let #u=e^x+1#, #=>#, #du=e^xdx#

Therefore,

#int(e^xdx)/(e^x+1)=int(du)/u=lnu=ln(e^x+1)+C#

So,

#int_(ln2)^(ln8)(e^xdx)/(e^x+1)=[ln(e^x+1)]_(ln2)^(ln8)#

#=(ln(e^(ln8)+1)-ln(e^(ln2)+1))#

#=ln(8+1)-ln(2+1)#

#=ln9-ln3#

#=ln(9/3)#

#=ln3#

#=1.099#